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So I've found that there's the Weierstrass Substitution that can be used on this problem but I just want to check I can use a normal substitution method to solve the equation:

$$\int \frac{\cos(x)}{\sin^2(x) +\sin (x)}dx$$

Let $u = \sin(x)$

$du = \cos(x)\, dx$

$dx = \frac {1}{\cos(x)\,} du$

Which becomes:

$$\int \frac{\cos(x)}{u^2 + u} \frac{1}{\cos(x)}du$$

$$\int \frac{1}{u^2 + u}du$$

Factor out u from denominator:

$$\int \frac{1}{u(u + 1)}du$$

Integrate as a partial fraction:

$$\int \frac{1}{u} - \frac{1}{(u + 1)}du$$

Which integrates as:

$$\ln|u| - \ln|(u + 1)| + C$$

Subtitute $u = \sin(x)$ back in and simplifies to:

$$\ln \left|\frac{\sin(x)}{\sin(x)+1} \right| + C$$

Is this correct? From the Weierstrass Substitution, one gets:

$$\ln \left|\tan \left(\frac{x}{2}\right)\right|-2\ln \left|\tan \left(\frac{x}{2}\right)+1\right| +C $$

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    $\begingroup$ Have you tried taking the derivative of your result? $\endgroup$ Commented Mar 4, 2016 at 19:59
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    $\begingroup$ $$u=\sin x\stackrel?\implies u+1=\sin(x+1)$$ I beg to differ... $\endgroup$
    – user228113
    Commented Mar 4, 2016 at 20:00
  • $\begingroup$ @G.Sassatelli Ah missed that, thank you for spotting that! $\endgroup$ Commented Mar 4, 2016 at 20:02
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    $\begingroup$ Note that you could start with a partial fraction approach right from the beginning and consider $$\int{\cos x\over \sin x}\text dx - \int{\cos x\over \sin x +1}\text dx$$ $\endgroup$
    – abiessu
    Commented Mar 4, 2016 at 20:24
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    $\begingroup$ @mostlyfabulous: you would have $u$-substitutions with $u = \sin x$ and $u=\sin x + 1$. Of course this approach simply brings about the exact same result as you found while having a few extra terms, so it's just another way to show the same thing. $\endgroup$
    – abiessu
    Commented Mar 4, 2016 at 20:31

3 Answers 3

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Manipulating the Weierstrass result, we begin with

$$\ln \left|\tan \left(\frac{x}{2}\right)\right|-2\ln \left|\tan \left(\frac{x}{2}\right)+1\right| +C$$

With $\tan\left(\frac x2\right)={\sin x\over 1+\cos x}$, we then get

$$\ln \left|{\sin x\over 1+\cos x}\right|-2\ln \left|{\sin x\over 1+\cos x}+1\right| +C\\ =\ln \left|(\sin x)(1+\cos x)\right|-2\ln \left|\sin x + \cos x+1\right| +C\\ =\ln \left|\sin x+\sin x\cos x\over(\sin x + \cos x+1)^2\right|+C\\ =\ln \left|\sin x+\sin x\cos x\over \sin^2 x+\cos^2 x +2\sin x\cos x+2\sin x+2\cos x+1\right|+C\\ =\ln \left|\sin x+\sin x\cos x\over 2 +2\sin x\cos x+2\sin x+2\cos x\right|+C_0\\ =\ln \left|\sin x(1+\cos x)\over (1+\sin x)(1+\cos x)\right|+C_1\\ =\ln \left|\sin x\over 1+\sin x\right|+C_1$$

So the two answers differ by a constant $(\ln 2)$.

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  • $\begingroup$ I don't quite follow how you manipulated the denominators here: $$\ln \left|(\sin x)(1+\cos x)\right|-2\ln \left|\sin x + \cos x+1\right| $$ $\endgroup$ Commented Mar 4, 2016 at 21:00
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    $\begingroup$ try to examine deeply and you may observe: $\ln|\sin{x}|-\ln|1+\cos{x}|-2\ln|\sin{x}+\cos{x}+1|+2\ln|1+\cos{x}|$ $\endgroup$
    – lcn
    Commented Mar 4, 2016 at 21:07
  • $\begingroup$ @mostlyfabulous: $$-2\ln|\sin x+\cos x +1| = \ln\left|\frac 1{(\sin x+\cos x+1)^2}\right|$$ $\endgroup$
    – abiessu
    Commented Mar 4, 2016 at 21:34
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$\int \frac{\cos x dx}{\sin^2 x+\sin x}=\int \frac{d \sin x}{\sin^2 x+\sin x}=\int\frac{dt}{t^2+t}=\ln \frac{t}{t+1}+ C=\ln \frac{\sin x}{\sin x+1}+C$

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  • $\begingroup$ But... this just restates what the comments have covered... $\endgroup$
    – abiessu
    Commented Mar 4, 2016 at 20:40
  • $\begingroup$ Sorry. I did not have read the comments $\endgroup$
    – Piquito
    Commented Mar 4, 2016 at 20:42
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By using Weierstrass Substitution,

let $t = \tan{\frac{x}{2}}, \,dt = \frac{1}{2}\sec^{2}{\frac{x}{2}}\,dx$

$=> 2\cos^{2}{\frac{x}{2}}\,dt = \frac{2}{1+t^{2}}\,dt = dx$

$\because \sin{x} = \frac{2t}{1+t^{2}}, \cos{x} = \frac{1-t^{2}}{1+t^{2}}$

$\therefore \int \frac{\cos{x}}{\sin^{2}{x}+\sin{x}}\,dx = \int \frac{(\frac{1-t^{2}}{1+t^{2}})(\frac{2}{1+t^{2}})}{(\frac{2t}{1+t^{2}})^2+(\frac{2t}{1+t^{2}})}\,dt = \int \frac{2(1-t^{2})}{4t^{2}+2t(1+t^{2})}\,dt = \int \frac{2(1-t)(1+t)}{2t(1+t^{2}+2t)}\,dt = \int \frac{1-t}{t(1+t)}\,dt$

$= \int \frac{1+t-2t}{t(1+t)} = \int \frac{1}{t}\,dt -2 \int \frac{1}{1+t}\,dt = \ln|t| - 2\ln|1+t| + C$

$ = \ln|\tan\frac{x}{2}|-2\ln|1+\tan\frac{x}{2}|+C$

Therefore your statement is correct!

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  • $\begingroup$ I saw this method and just want to know how I could be sure my statement is correct. That is: $$\ln \left|\frac{\sin(x)}{\sin(x)+1} \right| = \ln \left|\tan \left(\frac{x}{2}\right)\right|-2\ln \left|\tan \left(\frac{x}{2}\right)+1\right|$$ $\endgroup$ Commented Mar 4, 2016 at 20:50
  • $\begingroup$ I only want to show how to apply Weierstrass's method in this case. If you just want to know the identity, you may see the detailed abiessu's post. $\endgroup$
    – lcn
    Commented Mar 4, 2016 at 21:01

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